Download Physics 505 Fall 2007 Homework Assignment #1 — Solutions

Physics 505
Fall 2007
Homework Assignment #1 — Solutions
Textbook problems: Ch. 1: 1.5, 1.7, 1.11, 1.12
1.5 The time-averaged potential of a neutral hydrogen atom is given by
Φ=
αr q e−αr 1+
4π0 r
2
where q is the magnitude of the electronic charge, and α−1 = a0 /2, a0 being the Bohr
radius. Find the distribution of charge (both continuous and discrete) that will give
this potential and interpret your result physically.
We may obtain the charge distribution by computing ρ = −0 ∇2 Φ. However,
since Φ blows up as r → 0, we must be a bit careful. We first consider r > 0
q 1 ∂ 2 ∂
1 α
−αr
ρ = −0 ∇ Φ = −
r
e
+
4π r2 ∂r ∂r
r
2
2 2
q 1 ∂
α r
qα3 −αr
−αr
=
e
1
+
αr
+
=
−
e
4π r2 ∂r
2
8π
2
For r ≈ 0, on the other hand, we may expand
q
Φ=
4π0
1 α
− + ···
r
2
≈
q
4π0 r
This is the potential of a point charge q at the origin. Hence the complete charge
distribution can be written as
ρ = qδ 3 (r) −
qα3 −αr
e
8π
The first term corresponds to the proton charge, and the second to the negatively
charged electron cloud in the 1s orbital around the proton.
We can additionally verify that the hydrogen atom is indeed neutral
Z
Q=
qα3
ρd x = q −
8π
3
Z
0
∞
q
e−αr 4πr2 dr = q − Γ(3) = 0
2
1.7 Two long, cylindrical conductors of radii a1 and a2 are parallel and separated by a
distance d, which is large compared with either radius. Show that the capacitance per
unit length is given approximately by
−1
d
C = π0 ln
a
where a is the geometrical mean of the two radii.
This is essentially a two-dimensional electrostatic problem. We choose the geometry to be
a1
a2
d
x
To calculate the capacitance, we use C = Q/∆Φ where C is the capacitance per
unit length, Q (−Q) is the charge per unit length on the first (second) conductor,
and ∆Φ = Φ1 − Φ2 is the potential difference between the conductors.
The potential outside a single cylindrical conductor located at the origin is given
by the familiar expression
Q
Φ=−
log r
2π0
which may be obtained by integrating the electric field
~ = Q r̂
E
2π0 r
(which in turn is derived by straightforward application of Gauss’ law in integral
form). When the radii are small compared to the separation, we may superpose
the potentials for the two conductors to give
Φ(x) ≈ −
Q
[log x − log(d − x)]
2π0
which is valid for a1 ≤ x ≤ d − a2 (ie, for x between the two conductors). Note
that this is not an exact result, since the exact potential should be constant on
the surfaces of each conductor, while the above expression is not. The correction
terms are of order a1 /d or a2 /d, and may be dropped in this large separation
limit. We now compute the potential difference
Q
[− log a1 + log(d − a1 ) + log(d − a2 ) − log a2 ]
2π0
Q
(d − a1 )(d − a2 )
≈
log
2π0
a1 a2
2
Q
d
≈
log
2π0
a1 a2
∆Φ = Φ(a1 ) − Φ(d − a2 ) ≈
where we have assumed d a1 and d a2 . This gives the approximate expression for the capacitance
−1
−1
Q
d2
d
C=
≈ 2π0 log
= π0 log
∆Φ
a1 a2
a
where a =
√
(1)
a1 a2 is the geometric mean of the two radii.
Approximately what gauge wire (state diameter in millimeters) would be necessary
to make a two-wire transmission line with a capacitance of 1.2 × 10−11 F/m if the
separation of the wires was 0.5 cm? 1.5 cm? 5.0 cm?
This is one of the few problems where we will actually put in numbers. We first
rewrite (1) as
a = de−π0 /C
This show that, for a fixed capacitance per unit length, the relation between a
and d is linear. For the above value of the capacitance, we obtain
a ≈ 0.1d
Note that a is radius of the wire, while d is the separation. For the above numbers,
the wire diameters and gauges are approximately
Separation Diameter AWG Gauge Metric Gauge
0.5 cm
1 mm
18
10
1.5 cm
3 mm
9
30
5.0 cm
10 mm
OOO
100
Hopefully you agree that the metric wire gauges make much more sense.
1.11 Use Gauss’s theorem to prove that at the surface of a curved charged conductor, the
normal derivativce of the electric field is given by
1 ∂E
=−
E ∂n
1
1
+
R1
R2
where R1 and R2 are the principal radii of curvature of the surface.
We actually use Gauss’ law in integral form
Z
~ · n̂da = 0
E
S
when there are no charges enclosed. Before considering the three-dimensional
problem, consider the analogous situation in two dimensions
E
( R + ε) dθ
ε
R dθ
R
Take a curved Gaussian box next to the surface of the charged conductor at a
point where the radius of curvature is R. Gauss’ law then states
Z
~ · n̂da = Etop ∆atop − Ebottom ∆abottom
0=
E
(2)
S
where ∆atop and ∆abottom are the areas of the top and bottom of the box,
respectively. (There is no contribution from the sides of the box, because they are
taken to be normal to the surface.) Using ∆atop = (R + )dθdz and ∆abottom =
Rdθdz gives
0 = Etop (R + )dθdz − Ebottom Rdθdz
which yields the relation
Ebottom = Etop 1 +
R
This allows us to calculate
∂E
Etop − Ebottom
Etop
Etop
= lim
= lim −
=−
→0
→0
∂n
R
R
Noting that Etop is the same as E when → 0, this may be rewritten as
1 ∂E
1
=−
E ∂n
R
(3)
which is the analogous two-dimensional expression.
Coming back to the three-dimensional problem, we use the same method as above.
This time, however, the areas of the top and bottom of the Gaussian box are
∆atop = (R1 + )(R2 + )dΩ,
∆abottom = R1 R2 dΩ
Substituting this into (2) gives
Ebottom
= Etop 1 +
1+
R1
R2
which in turn yields
∂E
Etop − Ebottom
1
1
= lim
= lim −Etop
+
+
→0
→0
∂n
R1
R2
R1 R2
1
1
= −Etop
+
R1
R2
Rearranging this expression finally gives
1 ∂E
1
1
=−
+
E ∂n
R1
R2
(4)
Note that this reduces to the two-dimensional expression (3) in the cylindrical
limit R2 → ∞.
It is worth noting that, when proving the above, the Gaussian box is actually
taken to be infinitesimal. This suggests that we may use Gauss’ law in differential
form
~ ·E
~ =0
∇
(charge-free region)
Assuming the electric field is in the normal direction
~ = E n̂
E
then gives
~ ·E
~ =∇
~ · (E n̂) = n̂ · ∇E
~ + E∇
~ · n̂
0=∇
~ = ∂/∂n and rearranging gives
Using n̂ · ∇
1 ∂E
~ · n̂
= −∇
E ∂n
This is a neat expression because a bit of geometry indicates that the divergence of
the unit vector field is related to the principal radii of curvature of the associated
surface by
~ · n̂ = 1 + 1
∇
R1
R2
Substituting this into the above then reproduces (4).
Finally, note that neither of the derivations of (4) actually require the physical
existence of the conducting surface. Thus the expression (4) is valid in any chargefree region of space, provided we take E and n̂ to be the magnitude and direction
of the electric field. In this case, R1 and R2 represent the principal radii of
curvature of the equipotential surface corresponding to the electric field. With a
little bit of thought, it is easy to see that the physical interpretation of (4) is that
the electric field either gets weaker if the field lines diverge, or stronger if the field
lines converge. (Note that, formally, R1 and R2 can be positive or negative.)
1.12 Prove Green’s reciprocation theorem: If Φ is the potential due to a volume-charge
density ρ within a volume V and a surface-charge density σ on the conducting surface
S bounding the volume V , while Φ0 is the potential due to another charge distribution
ρ0 and σ 0 , then
Z
Z
Z
Z
0 3
0
0
3
ρΦ d x +
σΦ da =
ρ Φd x +
σ 0 Φda
V
S
V
S
We start with Green’s theorem for the potentials Φ and Φ0
Z
0
2
0
2
Z 3
(Φ∇ Φ − Φ ∇ Φ)d x =
V
S
∂Φ0
∂Φ
Φ
− Φ0
∂n
∂n
da
On the left hand side, we use
∇2 Φ = −
1
ρ
0
∇ 2 Φ0 = −
This gives
Z
0
0
Z 3
(ρΦ − ρ Φ)d x = 0
V
S
Z
∂Φ0
∂Φ
− Φ0
Φ
∂n
∂n
0
(Φ E⊥ −
= 0
1 0
ρ
0
da
(5)
0
ΦE⊥
)da
S
where E⊥ is the normal electric field at the surface of the conducting surface,
~ = −∂Φ/∂n. Since n̂ is an outward pointing normal, and since E
~ is
E⊥ ≡ n̂ · E
the electric field on the interior of the surface, application of Gauss’ law at the
conducting surface gives
1
E⊥ = − σ
0
Substituting this into (5) and rearranging then gives
Z
0 3
Z
ρΦ d x +
V
0
Z
σΦ da =
S
0
3
Z
ρ Φd x +
V
S
σ 0 Φda